Linearly boundedness implies boundedness of solution

Question

Let $(X,\vert\vert\cdot\vert\vert)$ be a Banach space and $f:\mathbb R \times X \to X$ continuous and linearly bounded in its second variable, i.e. there are functions $\alpha, \beta \in C(\mathbb R; \mathbb R_0^+) \cap L^1(\mathbb R)$ such that $$\vert\vert f(t,v)\vert\vert \leq \alpha(t)+\beta(t)\vert\vert v\vert\vert$$ for any $(t,v) \in \mathbb R \times X$.

How can I show that every solution of $$u'(t)=f(t,u(t)), u(t_o)=u_o \in X$$$t,t_0\in \mathbb R$ is bounded?

Answer 1

We have for $t>t_0$ $$\begin{align} \|u(t)\|=&\Bigl\|\,u(t_0)+\int_{t_0}^tf(s,u(s))\,ds\,\Bigr\|\\ &\le\|u(t_0)\|+\int_{t_0}^t\bigl(\alpha(s)+\beta(s)\,\|u(s)\|\bigr)\,ds\\ &\le \|u(t_0)\|+\int_{t_0}^\infty\alpha(s)\,ds+\int_{t_0}^t\beta(s)\,\|u(s)\|\,ds. \end{align}$$ Now use Gronwalls's lemma.

The same argument works for $t<t_0$.

Answer 2

First for Hilbert space, let $x(t)=\|u(t)\|^2$. Then \begin{eqnarray} x'(t)&=&2(u'(t),u(t))=2(f(t,u(t)),u(t))\\ &\le&\|f(t,u(t))\|\|u(t)\|\\ &\le&2\alpha(t)\|u(t)\|+2\beta(t)x(t)\\ &\le& \alpha(t)(x(t)+1)+2\beta(t)x(t)\\ &=&\alpha(t)+(\alpha(t)+2\beta(t))x(t) \end{eqnarray} from which we have $$ (x(t)e^{-\int_0^t(\alpha(\tau)+2\beta(\tau))d\tau})'\le \alpha(t)e^{-\int_0^t(\alpha(\tau)+2\beta(\tau))d\tau}. $$ Integrating both sides from $0$ to $t$, one has $$ x(t)\le x(0)+ \int_0^t\alpha(s)e^{\int_s^t(\alpha(\tau)+2\beta(\tau))d\tau}ds.$$ Since $\alpha,\beta(t)\in L^1(R)$, it is easy to see $x(t)<M$ for some $M>0$ for $\forall t\ge 0$ and hence $u(t)$ is bounded. For Banach space, the situation is almost identical.

Answer 3

You get approximately up to quadratic terms in $h$ \begin{align} \|u(t+h)\|&\le \|u(t)\|+h·\|f(t,u(t))\| \\ &\le\|u(t)\|+h·α(t)+h·β(t)||u(t)|| \\ &=(1+h·β(t))\|u(t)\|+h·α(t) \le e^{h·β(t)}\|u(t)\|+h·α(t) \end{align} In the next step you get $$ \|u(t+2h)\|\le e^{h·β(t+h)}\left(e^{h·β(t)}\|u(t)\|+h·α(t)\right)+h·α(t+h) $$ so that in continuation this becomes $$ \|u(t+kh)\|\le e^{h·\sum_{j=0}^{k-1}β(t+jh)}\|u(t)\|+h·\sum_{m=0}^{k-1}α(t+mh)e^{h·\sum_{j=m+1}^{k-1}β(t+jh)} $$ which obviously transfers in the limit $h\to0$ into integrals $$ \|u(t_1)\| \le e^{\int_{t_0}^{t_1}β(s)\,ds}\|u(t_0)\| +\int_{t_0}^{t_1}α(r)e^{\int_{r}^{t_1}β(s)\,ds}dr $$ Now using that $ α$ and $β$ are $L^1$ integrable one easily gets an upper bound for $u$. Using the monotonously falling functions $A(t)=\int_t^{\infty}α(s)\,ds$ and $B(t)=\int_t^{\infty}β(s)\,ds$ one gets \begin{align} \|u(t_1)\| &\le e^{-B(t_1)}\left(e^{B(t_0)}\|u(t_0)\| + \int_{t_0}^{t_1}α(r)e^{B(r)}dr\right)\\ &\le e^{B(t_0)}\left(\|u(t_0)\|+A(t_0)\right) \end{align}